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umka2103 [35]
2 years ago
12

Suppose the foreman had released the box from rest at a height of 0.25 m above the ground. What would the crate's speed be when

it reaches the bottom of the ramp
Physics
1 answer:
Arturiano [62]2 years ago
4 0

Answer:

v = 2.21 m/s

Explanation:

The foreman had released the box from rest at a height of 0.25 m above the ground.

We need to find the speed of the crate when it reaches the bottom of the ramp. Let v is the velocity at the bottom of the ramp. It can be calculated using conservation of energy as follows :

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 0.25} \\\\v=2.21\ m/s

So, its velocity at the bottom of the ramp is 2.21 m/s.

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A mine elevator is supported by a single steel cable 0.0125 m in diameter. The total mass of the elevator cage and occupants is
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0.8895m

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L = 362m

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